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Solution
Your input: simplify and calculate different forms of $$$81 i$$$
The expression is already simplified.
Polar form
For a complex number $$$a+bi$$$, polar form is given by $$$r(\cos(\theta)+i \sin(\theta))$$$, where $$$r=\sqrt{a^2+b^2}$$$ and $$$\theta=\operatorname{atan}\left(\frac{b}{a}\right)$$$
We have that $$$a=0$$$ and $$$b=81$$$
Thus, $$$r=\sqrt{\left(0\right)^2+\left(81\right)^2}=81$$$
Also, $$$\theta=\operatorname{atan}\left(\frac{81}{0}\right)=\frac{\pi}{2}$$$
Therefore, $$$81 i=81 \cos{\left(\frac{\pi}{2} \right)} + 81 i \sin{\left(\frac{\pi}{2} \right)}$$$
Inverse
The inverse of $$$81 i$$$ is $$$\frac{1}{81 i}$$$
Multiply and divide by $$$i$$$ (keep in mind that $$$i^2=-1$$$):
$$${\color{red}{\left(\frac{1}{81 i}\right)}}={\color{red}{\left(- \frac{i}{81}\right)}}$$$
Hence, $$$\frac{1}{81 i}=- \frac{i}{81}$$$
Conjugate
The conjugate of $$$a + i b$$$ is $$$a - i b$$$: the conjugate of $$$81 i$$$ is $$$- 81 i$$$
Modulus
The modulus of $$$a + i b$$$ is $$$\sqrt{a^{2} + b^{2}}$$$: the modulus of $$$81 i$$$ is $$$81$$$
Answer
$$$81 i=81 i=81.0 i$$$
The polar form of $$$81 i$$$ is $$$81 \cos{\left(\frac{\pi}{2} \right)} + 81 i \sin{\left(\frac{\pi}{2} \right)}$$$
The inverse of $$$81 i$$$ is $$$\frac{1}{81 i}=- \frac{i}{81}\approx - 0.0123456790123457 i$$$
The conjugate of $$$81 i$$$ is $$$- 81 i=- 81.0 i$$$
The modulus of $$$81 i$$$ is $$$81$$$